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Now I will apply Newton's laws of motion and gravity to topics more astronomical:
objects moving around other objects. What kinds of things can you find out about
celestial objects from just observing their motions?
Newton's first law of motion says that an object's inertia will keep it
from changing its speed and/or direction unless some force acts on it. This
means that satellites orbiting the Earth must be feeling some force that
constantly deflects them toward the center of the Earth. If there was no force,
they would travel in a straight line at a constant speed.
If you whirl a ball attached to string around your head, it moves in a circular
path around you because the string is always pulling the ball directly toward
the hand grabbing the string. The ball wants to move in a straight line and the
string is pulling it directly inward. The resulting deflection is a compromise:
a circular path. The string is applying a centripetal force to the ball:
an inward force. If you let go of the string, there is no centripetal force and
the ball will fly off in a straight line because of its inertia. If you do not
whirl the ball fast enough it will move inward to a smaller non-circular path
around you. If you whirl the ball too fast, you may not be able to give it
enough centripetal force to keep it in a circular path around you.
The amount of centripetal force needed to balance an object's inertia and keep
it in a circular path of radius r is found from Newton's second law:
the centripetal force = m v2 / r, where v and m are
the object's speed and mass, respectively. The radius of the orbit r is the same
as the distance between the moving object and the central body.
Now for orbits! Satellites are not being deflected by strings but by gravity.
Gravity provides the centripetal force needed to keep the satellites in orbit.
If you focus on the simple case of circular orbits, you can use the centripetal
force formula above with the law of gravity to determine the mass of a
planet or star. Simply set the force of gravity equal to
the centripetal force and solve for the mass of the planet or star, M.
The satellite mass m cancels out from both sides and if you put M on one
side and the rest on the other side of the equation, you get
This assumes that the satellite's mass, m, is much less than the
central object's mass so you can ignore the acceleration of the central object
toward the satellite!
How do you do that?
Let's use this result to get an estimate of the mass of the Sun. You need to use
something orbiting with a known radius and speed. The Earth's orbit
is roughly circular with radius = 1.5 × 1011
meters and the Earth moves with a speed 30,000 meters/second (= 30 km/s) in its orbit.
The distance is given in meters to match the units of the speed. The distance unit of
a meter is used because you will be using the gravitational constant G in your
calculation and it uses the meter unit. When you do a calculation, you must be sure you
check that your units match up or you will get nonsense answers.
Plug the values into the mass relation:
the Sun's mass = (30,000)2 × (1.5 ×
1011)/(6.7 × 10-11) = 2 × 1030 kilograms.
This is much larger than the
Earth's mass so it was okay to ignore the Sun's movement toward the Earth.
Using no approximations (ie., not assuming a circular orbit and including the
Sun's motion toward the Earth) gives a value for the Sun's mass that is very
close to this. Your answer does not depend on which planet you choose to use (here
you used the Earth's orbit). You would get the same value for the mass of the Sun
if you had used any of the other planets orbital speeds and sizes.
|
This relation tells you what you need to know in order to measure a planet's or
star's mass:
the orbital speed of a satellite and the distance it is from the center of the
planet or star. Because the velocity is on top of the fraction, satellites
are made to move faster if the mass of the central object is greater. At the
same distance, a massive
planet will exert more gravity force than a low-mass planet, so the
massive planet will produce greater inward accelerations on satellites orbiting it.
The satellites will, therefore, orbit at faster speeds.
Sometimes the orbital period P is measured
instead of the orbital velocity. The orbital period is the time it takes the object to
travel the circumference of its orbit (for a circle, the circumference =
2pr, where
p is
approximately 3.1416). Recall that speed = (distance travelled)/time,
so the speed v = the circumference of the orbit/orbital period. When you
substitute this for the speed in the mass relation above, you get
This may look familiar to you---it is Kepler's third law! There is a distance
cubed, an orbital period squared, and some other contant factors. Newton found
that when Kepler used the motions of the planets to formulate his third law, Kepler
was actually measuring the mass of the Sun. If you use the convenient set of
units of an astronomical unit for the distance, a year for the time, and a
``solar mass'' (mass relative to the Sun) for the mass, the complicated term
(4p2)/G becomes the simple value of 1. For
the planets orbiting the Sun, the mass relation becomes 1 = r3 /
P2, or r3 = P2, just
what Kepler found.
The mass formula above tells you that satellites orbiting massive planets must
move faster than satellites orbiting low-mass planets at the same distance.
Massive planets have stronger gravity than low-mass planets so a satellite
orbiting a massive planet is accelerated by a greater amount than one going
around a lesser mass planet at the same distance. To balance the stronger inward
gravitational pull of the massive planet, the satellite must move faster in its
orbit than if it was orbiting a lesser mass planet. Of course, this also applies
to planets orbiting stars, stars orbiting other stars, etc.
If you solve for the orbit speed, v, in the mass formula, you
can find how fast something needs to move to balance the inward pull of gravity:
v2 = (G M)/r .
Taking the square root of both sides (you want just v not
v2), you get
v = Sqrt[(G M)/r].
How do you do that?
Find the orbital speed of Jupiter around the Sun. Jupiter's distance from
the Sun is 5.2 A.U., or 7.8×1011 meters and the Sun's mass is
2×1030 kilograms.
The orbital speed of Jupiter around the Sun is
Sqrt[6.7×10-11 ×
(2×1030)/(7.8×1011)] =
Sqrt[1.718×108] = 1.3×104 meters/second,
or 13 kilometers/second. What do you think you wouldd find if you used one of the Trojan
asteroids, millions of times less massive than Jupiter, that orbits the Sun at 5.2 A.U.?
|
If an object moves fast enough it can escape a massive object's gravity and
not be drawn back toward the massive object. The critical speed needed to do
this is the escape velocity. More specifically, this is the initial
speed something needs to escape the object's gravity and assumes that there is no
other force acting on the
object besides gravity after the initial boost. Rockets leaving the Earth do
not have the escape velocity at the beginning but the engines provide thrust
for an extended period of time, so the rockets can eventually escape. The concept
of escape velocity applies to anything
gravitationally attracted to anything else (gas particles in planet
atmospheres, comets orbiting the Sun, light trying to escape from black holes,
galaxies orbiting each other, etc.).
Using Newton's laws of motion and law of gravity, you can find that the escape
velocity vesc looks very similar to the orbital speed:
vesc = Sqrt[(2 G M)/r].
This is a factor Sqrt[2] larger than the
circular orbital speed.
Since the mass M is on top of the fraction, the escape velocity
increases as the mass increases. More massive bodies exert greater gravity
force, so escaping objects have to move faster to overcome the greater gravity.
Also, the distance from the center of the object r is in the bottom of the
fraction, so the escape velocity DEcreases as the distance increases. Gravity
decreases with greater distance, so objects farther from a massive body do not need to
move as quickly to escape it than those closer to it.
How do you do that?
Find the escape velocity from the surface of the Earth.
Using the
acceleration of gravity, you can find that the Earth has a mass
of 6.0×1024 kilograms. The Earth's radius is 6.4×106
meters. Since the mass and distance from the center are in the standard units, you just
need to plug their values into the escape velocity relation.
The Earth's surface escape velocity is
Sqrt[2× (6.7×10-11) × (6.0×1024)/
(6.4×106)] = Sqrt[1.256×108] =
1.1×104 meters/second (= 11 km/s). Here are some other surface
escape velocities: Moon = 2.4 km/s, Jupiter = 59.6 km/s, Sun = 618 km/s.
|
Vocabulary
| centripetal force | escape velocity |
Formulae
- Mass of central object = [(orbital speed)2 × distance)/G.
- Mass of central object (Kepler's 3rd law) = (4p2)/G
× [(distance)3/(orbital period)2].
- Orbital speed = Sqrt[G × Mass / distance].
- Escape velocity = Sqrt[2G × Mass / distance].
- What keeps satellites orbiting the Earth moving along their curved paths?
- What two things must be determined first in order to calculate the mass of a
planet or a star?
- Jupiter's moon Io has about the same mass as the Moon and orbits Jupiter at
about the same distance that the Moon orbits the Earth (center to center). Then
why does Io take only 1.8 days to orbit Jupiter but our Moon takes 27.3 days to
orbit the Earth?
- Astronomers were able to accurately measure the orbital periods of the moons
of Jupiter since the time of Galileo, so why was an accurate value for Jupiter's
mass not found for over 300 years until the astronomical unit was measured
accurately?
- Which would have a shorter orbital period, a planet orbiting a massive
star at 3 A.U. or a planet orbiting a low-mass star at 3 A.U.? Explain your answer.
- If a planet orbiting a massive star has the same orbital period as a
planet orbiting a low-mass star, which of the planets orbits at a greater distance
from its star? Explain your answer.
- What two things does the escape velocity depend on?
- Why does the planet Saturn with over 95 times the Earth's mass have a
smaller escape velocity at its cloudtops than the Earth has at its cloudtops?
- Why is Jupiter's escape velocity at its
cloudtops over two times higher than the Earth's surface escape
velocity, even
though Jupiter has a much larger diameter than the Earth?
Kepler's third law of planetary motion says that the average distance of a
planet from the Sun cubed is directly proportional to the orbital
period squared. Newton found that his gravity force law could explain Kepler's
laws. Since Newton's law of gravity applies to any object with mass, Kepler's
laws can be used for any object orbiting another object. Let's look at
satellites orbiting a planet.
If you have two satellites (#1 and #2) orbiting a planet, Kepler's third law
says:
(period #1/period #2)2 = (distance #1/distance #2)3,
where the distance is the average distance of the
satellite from the planet---the orbit's semimajor axis. The satellites must be
orbiting the same planet in order to use Kepler's third law! Kepler found this law
worked for the planets because they all orbit the same star (the Sun).
If you have measured the orbital period of
one satellite around a planet, you can then easily find how long it would take
any other satellite to orbit the planet in any size oribt. Kepler's third law can
be simplified down to
| period #1 | = |
period #2 × Sqrt[(distance #1/distance #2)3] |
| OR |
|
| period #1 | = |
period #2 × (distance #1/distance #2)3/2. |
Those of you with a scientific calculator (one that does powers, trig functions,
scientific notation, etc.) will want to use the formula on the last line
(remember that 3/2 = 1.5). Those
with a calculator that just has a square root button will want to use the
formula on the second-to-last line.
If the satellite is orbiting the Sun,
then the relation can be greatly simplified with an appropriate choice of
units: the unit of years for the orbit period
and the distance unit of astronomical units. In this case, the
reference ``satellite'' is the Earth and Kepler's third law becomes
period = distance3/2. Let's use this to find out how long it takes
to explore the solar system.
The simplest way to travel between the planets is to let the Sun's gravity do
the work and take advantage of Kepler's laws of orbital motion. A fuel efficient
way to travel is to put the spacecraft in orbit around the Sun with the Earth
at one end of the orbit at launch and the other planet at the opposite end at
arrival. These orbits are called ``Hohmann orbits'' after Walter Hohmann who
developed the theory for transfer orbits. The spacecraft requires only an
acceleration at the beginning of the trip and a deceleration at the end of the
trip to put it in orbit around the other planet.
Let's
go to Mars! The relative positions of Earth and
Mars must be just right
at launch so that Mars will be at the right position to greet the spacecraft
when it arrives several months later. These good positionings happen once every
780 days (the synodic period of Mars). The spacecraft must be launched within a
time interval called the ``launch window'' that is just few of weeks long to use a
Hohmann orbit for the spacecraft's path. The Earth is at the
perihelion (point closest to the Sun) of the spacecraft orbit (here,
1.0 A.U.) and Mars is at the aphelion (point farthest from the Sun---here,
1.52 A.U.).
Kepler's third law relates the semi-major axis of the orbit to its sidereal
period. The major axis is the total length of the long axis of the elliptical
orbit (from perihelion to aphelion). For the Mars journey, the major axis =
1.52 + 1.0 A.U. = 2.52 A.U. The semi-major axis is one-half of the major
axis, so divide the major axis by two: 2.52/2 = 1.26 A.U. Now apply Kepler's
third law to find the orbital period of the spacecraft = 1.263/2 =
1.41 years. This is the period for a full orbit (Earth to Mars and back to Earth),
but you want to go only half-way (just Earth to Mars). Travelling from Earth to
Mars along this path will take (1.41 / 2) years = 0.71 years or about 8.5 months.
When the craft is launched, it already has the Earth's orbital velocity of
about 30 km/sec. Since this is the speed for a circular orbit around the Sun
at 1.0 A.U., a reduction in the spacecraft's speed would make it fall closer
to the Sun and the Hohmann would be inside the Earth's orbit. Since you
want to go beyond the Earth's orbit, the spacecraft needs an increase in its
speed to put it in an orbit that is outside the Earth's orbit. It will slow
down gradually as it nears aphelion.
At aphelion the spacecraft will not be
travelling fast enough to be in a circular orbit at Mars' distance (1.52 A.U.)
so it will need to arrive at aphelion slightly before Mars does. Mars
will then catch up to it. But the spacecraft will be moving much too fast to be in
a circular orbit around Mars, so it will need to slow down to go in orbit
around Mars.
On its journey to Mars, the spacecraft's distance from the Sun is continuously
monitored to be sure the craft is on the correct orbit. Though the spacecraft
responds mostly to the Sun's gravity, the nine planets' gravitational pulls on the
spacecraft can
affect the spacecraft's path as it travels to Mars, so occasional minor firings
of on-board thrusters may be required to keep the craft exactly on track.
Vocabulary
Formulae
- Kepler's third law: period #1 = period #2 ×
Sqrt[(distance #1/distance #2)3]
- Kepler's third law: period #1 = period #2 ×
(distance #1/distance #2)3/2
- If considering objects orbiting the Sun, measure the orbit period in years
and the distance in A.U. With these units, Kepler's third law is simply:
period = distance3/2.
- How can you predict the orbital period of Jupiter's satellite Europa from
observations of the other jovian moon Io?
- If Io takes 1.8 days to orbit Jupiter at a distance of 422,000 kilometers
from its center, find out how long it would take Europa to orbit Jupiter at
671,000 kilometers from its center.
- If the Moon were twice as far from the Earth as it is now, how long could
a solar eclipse last? (Solar eclipses currently last up to about two hours from
the start of the cover-up to when the Moon no longer blocks the Sun at all.)
- The Hubble Space Telescope orbits the Earth 220 kilometers above the surface
and takes about 1.5 hours to complete one orbit. How can you find out how far
up to put a communication satellite, so that it takes 24 hours to circle the Earth?
(Such an orbit is called a ``geosychronous orbit'' because the satellite remains
above a fixed point on the Earth.)
- Why does NASA not launch interplanetary spacecraft when the planets are at
opposition (closest to the Earth)?
- Find out how long it will take the Cassini spacecraft to travel to Saturn
9.5 A.U. from the Sun.
When you look in the paper at the section containing the tide tables, you will
often see the phase of the moon indicated as well. That is because the ocean
tides are
caused by different strengths of the Moon's gravity at different points on the
Earth.
The side of the Earth facing the Moon
is about 6400 kilometers closer to the Moon than the center of the Earth is, and the
Moon's gravity pulls on the near side of the Earth more strongly than on the
Earth's
center. This produces a tidal bulge on the side of the Earth facing the Moon.
The
Earth rock is not perfectly rigid; the side facing the Moon responds by rising
toward the Moon by a few centimeters on the near side. The more fluid seawater
responds by flowing into a bulge on the side of the Earth facing the Moon. That
bulge is the high tide.
At the same time the Moon exerts an attractive force on
the
Earth's center that is stronger than that exerted on the side away from the
Moon.
The Moon pulls the Earth away from the oceans on the far side, which flow into
a bulge
on the far side, producing a second high tide on the far side.
These tidal
bulges
are always along the Earth-Moon line and the Earth rotates beneath the tidal
bulge.
When the part of the Earth where you are located sweeps under the bulges, you
experience a high tide; when it passes under one of the depressions, you experience
a low tide. An ideal coast should experience the rise and fall of the tides twice
a day. In reality, the tidal cycle also depends on the latitude of the site, the
shape of the shore, winds, etc.
The Sun's gravity also produces tides that are about half as strong as the
Moon's
and produces its own pair of tidal bulges. They combine with the lunar tides. At
new
and full moon, the Sun and Moon produce tidal bulges that add together to
produce
extreme tides. These are called spring tides (the waters really spring
up!). When the Moon and Sun are at right angles to each other (1st & 3rd quarter),
the solar tides reduce the lunar tides and you have neap tides.
As the Earth rotates beneath the tidal bulges, it attempts to drag the bulges
along with it. A large amount of friction is produced which slows down the
Earth's spin.
The day has been getting longer and longer by about 0.0016 seconds each century.
Over the course of time this friction can have a noticeable effect. Astronomers
trying
to compare ancient solar eclipse records with their predictions found that they
were
off by a significant amount. But when they took the slowing down of the Earth's
rotation into account, their predictions agreed with the solar eclipse records.
Also, growth rings in ancient corals about 400 hundred million years old show
that the day
was only 22 hours long so that there were over 400 days in a year. In July
1996 a research study reported evidence, from several sedimentary rock records
providing an indicator of tidal periods, that the day was only 18 hours long
900 million years ago.
Eventually the
Earth's rotation will slow down to where it keeps only one face toward the Moon.
Gravity acts both ways so the Earth has been creating tidal bulges on the Moon
and
has slowed it's rotation down so much that it rotates once every orbital period.
The Moon keeps one face always toward the Earth.
Here is a list of references about the evidence
for the slowing down of the Earth's rotation:
- Growth Rhythms and the History of the Earth's rotation, edited by
G.D. Rosenberg and S.K. Runcorn (Wiley: New York, 1975). An excellent source on
the eclipse records and the biology of coral and their use as chronometers.
- Tidal Friction and the Earth's Rotation, edited by P. Brosche and
J. S¸ndermann (Springer Verlag, 1978). The second volume put out in 1982
does not talk about eclipse records or the use of coral but, instead, goes into
the astrophysics of the Earth-Moon dynamics and geophysics of internal Earth
processes effects on the Earth's rotation.
- Earth's Rotation from Eons to Days, edited by P. Brosche and J.
S¸ndermann (Springer Verlag, 1990). Has several articles about the use of
ancient Chinese observations.
- Richard Monastersky 1994, Ancient tidal fossils unlock lunar secrets
in Science News vol. 146, no. 11, p. 165 of the 10 Sept 1994 issue.
- C. P. Sonett, E. P. Kvale, A. Zakharian, Marjorie A. Chan, T. M. Demko
1996,
Late
Proterozoic and Paleozoic Tides, Retreat of the Moon, and
Rotation of the Earth in Science vol 273, no. 5271, p. 100 of
the 05 July 1996 issue.
Friction with the ocean beds drags the tidal bulges eastward out of a direct
Earth-Moon line and since these bulges contain a lot of mass, their gravity
pulls the moon forward in its orbit. The increase in speed enlarges the
Moon's orbit. Currently, the Moon's distance from the Earth is increasing by
about 3 centimeters per year. Astronomers have been able to measure this slow
spiralling out of the Moon by bouncing laser beams off reflectors left by
the Apollo astronauts on the lunar surface.
The consequence of the Moon's recession from the Earth because of the slowing
down
of the Earth's rotation is also an example of the conservation of angular
momentum. Angular momentum is the amount of spin motion an object or group of
objects has. It depends on the geometric size of the object or group of objects,
how fast the object (or group of objects) is moving, and the mass of the object
(or the group). Since the Earth's angular momentum is decreasing, the Moon's angular
momentum must increase to keep the overall angular momentum of the
Earth-Moon system the same.
The concept of angular momentum is discussed further in the
Angular Momentum appendix.
The slow spiralling out of
the Moon means that there will come a time in the future when the angular size
of
the Moon will be smaller than the Sun's and we will not have any more total solar
eclipses! Fifty billion years in the future the Earth day will equal 47 of our
current days and the Moon will take 47 of our current days to orbit the Earth.
Both will be locked with only one side facing the other---people on one side
of the Earth will always see the Moon while people on the other side will only
have legends about the Moon that left their pleasant sky.
Tidal effects are larger for more massive objects and at closer distances. The
Sun produces a tidal bulge on the planet Mercury (the planet closest to the Sun)
and has slowed that planet's rotation period so it rotates three times for every
two times it orbits the Sun (a ``3 - 2 spin-orbit resonance''). Jupiter's moon, Io,
orbits at about the same distance from Jupiter's center as the Earth's moon. Jupiter
is much more massive than the Earth, so Jupiter's tidal effect on Io is much greater
than the Earth's tidal effect on the Moon. Io is stretched by varying amounts as it
orbits Jupiter in its elliptical orbit. This tidal flexing of the rock material
creates huge amounts of heat from friction in Io's interior which in turn is
released in many volcanic eruptions seen on Io. Galaxies passing close to each other
can be severely stretched and sometimes pulled apart by mutual tidal effects.
Vocabulary
| conservation of angular momentum | neap tide | spring tide |
- What causes the tides?
- How are tides related to the position of the Moon and Sun with respect to
the Earth?
- Why are there two high tides roughly every 12.5 hours? Explain why
there are two tidal bulges AND why they are over 12 hours apart.
- At what phases do spring tides occur?
- At what phases do neap tides occur?
- How are tides responsible for the slowing down of the Earth's spin and
the Moon's spiralling away from us?
- Where are some other places that tides play a significant role in the
appearance and motion of objects?
Go to
Gravity Introduction section
last updated 25 January 1999
Nick Strobel --
Email:
strobel@lightspeed.net
(661) 395-4526
Bakersfield College
Physical Science Dept.
1801 Panorama Drive
Bakersfield, CA 93305-1219
